NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2
NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2: Academic team of PW uploaded errorless NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 prepared based on recommendations of CBSE.
Krati Saraswat9 Feb, 2024
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NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 (Applications of Integrals)
NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 Applications of Integrals is prepared by the academic team of Physics Wallah. We have prepared NCERT Solutions for all exercises of Chapter 8. Given below is step by step solutions of all questions given in NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2.NCERT Solutions for Class 12 Maths Chapter 8 Miscellaneous Exercise
NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 Overview
NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 covers these important topics. Students are encouraged to review each topic thoroughly in order to fully understand the concepts taught in the chapter and make optimal use of the provided solutions. These solutions are the outcome of the dedicated effort that the Physics Wallah teachers have been doing to aid students in understanding the ideas covered in this chapter. After going over and rehearsing these responses, the goal is for students to easily score outstanding exam results.NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2
Solve The Following Questions NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2:
Question 1. Find the area of the circle 4 x 2 + 4 y 2 = 9 which is interior to the parabola x 2 = 4 y Solution : The required area is represented by the shaded area OBCDO.
Solving the given equation of circle, 4
x
2
+ 4
y
2
= 9, and parabola,
x
2
= 4
y
, we obtain the point of intersection as B(√2,1/2) and D (-√2,1/2).
It can be observed that the required area is symmetrical about
y
-axis.
∴ Area OBCDO = 2 × Area OBCO
We draw BM perpendicular to OA.
Therefore, the coordinates of M are (√2,0).
Therefore, Area OBCO = Area OMBCO – Area OMBO
Therefore, the required area OBCDO is
NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.1
Question 2. Find the area bounded by curves ( x – 1) 2 + y 2 = 1 and x 2 + y 2 = 1 Solution : The area bounded by the curves, ( x – 1) 2 + y 2 = 1 and x 2 + y 2 = 1, is represented by the shaded area as
On solving the equations, (
x
– 1)
2
+
y
2
= 1 and
x
2
+
y
2
= 1, we obtain the point of intersection as A (1/2,√3/2) and B (1/2,√3/2) .
It can be observed that the required area is symmetrical about
x
-axis.
∴ Area OBCAO = 2 × Area OCAO
We join AB, which intersects OC at M, such that AM is perpendicular to OC.
The coordinates of M are (1/2,0).
Therefore, required area OBCAO =
units.
Question 3. Find the area of the region bounded by the curves
y
=
x
2
+ 2,
y
=
x
,
x
= 0 and
x
= 3
Solution :
The area bounded by the curves,
y
=
x
2
+ 2,
y
=
x
,
x
= 0, and
x
= 3, is represented by the shaded area OCBAO as
Then, Area OCBAO = Area ODBAO – Area ODCO
Question 4. Using integration, find the area of the region bounded by the triangle whose vertices are (–1, 0), (1, 3) and (3, 2).
Solution :
BL and CM are drawn perpendicular to
x
-axis.
It can be observed in the following figure that,
Area (ΔACB) = Area (ALBA) + Area (BLMCB) – Area (AMCA) … (1)
Therefore, from equation (1), we obtain
Area (ΔABC) = (3 + 5 – 4) = 4 units.
5. Using integration, find the area of the triangular region whose sides have the equations
y
= 2
x
+1,
y
= 3
x
+ 1 and
x
= 4.
Solution :
The equations of sides of the triangle are
y
= 2
x
+1,
y
= 3
x
+ 1, and
x
= 4.
On solving these equations, we obtain the vertices of triangle as A(0, 1), B(4, 13), and C (4, 9).
It can be observed that,
Area (ΔACB) = Area (OLBAO) –Area (OLCAO)
Question 6. Choose the correct answer:
Smaller area enclosed by the circle
x
2
+
y
2
= 4 and the line
x
+
y
= 2 is
(A)2 (π – 2)
(B). π – 2
(C). 2π – 1
(D). 2 (π + 2)
Solution : The smaller area enclosed by the circle, x 2 + y 2 = 4, and the line, x + y = 2, is represented by the shaded area ACBA as
It can be observed that,
Area ACBA = Area OACBO – Area (ΔOAB)
Therefore, option (B) is correct.
Question 7. Choose the correct answer:
Area lying between the curves
y
2
= 4
x
and
y
= 2
x
is
(A) 2/3
(B) 1/3
(C) 1/4
(D) 3/4
Solution :
The area lying between the curve,
y
2
= 4
x
and
y
= 2
x
, is represented by the shaded area OBAO as
The points of intersection of these curves are O (0, 0) and A (1, 2).
We draw AC perpendicular to
x
-axis such that the coordinates of C are (1, 0).
∴ Area OBAO = Area (OCABO) – Area (ΔOCA)
Therefore, option (B) is correct.
NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 FAQs
Where can I find NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2?
NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 are available on Physics Wallah.
Why are NCERT Solutions important for Class 12 Maths Chapter 8 Exercise 8.2?
NCERT Solutions provide step-by-step explanations and solutions to the questions given in the textbook. They help students understand the concepts better, clarify doubts, and practice solving problems effectively. NCERT Solutions are aligned with the CBSE curriculum, making them essential for exam preparation.
How should I use NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 effectively?
To use NCERT Solutions effectively, students should first attempt to solve the questions on their own. After attempting the questions, they can refer to the NCERT Solutions to verify their answers, understand the solution approach, and learn alternate methods of solving problems. Regular practice with NCERT Solutions enhances problem-solving skills and boosts confidence.
Are NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 helpful for competitive exams?
Yes, NCERT Solutions for Class 12 Maths Chapter 8 Exercise 8.2 are helpful for various competitive exams, especially those that have a syllabus aligned with the CBSE curriculum.
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